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2.Вычислите f’(П/3),если
f’(x)=2sinx +3x2 -2Пx+3= 2*(<span>sqrt{3}/2) + 3*((П^2)/9 - 2(П^2)/3 + 3</span>
3.найдите производную
y=4/x
y'=-4/(x^2)
y=2√x+3 sinx
y'=1/2√x + 3*(cosx)
y=3-2x
y'= -2
Лови, вроде должно быть так
{(x-2)y=6 => y=6/(x-2) ОДЗ: x≠2<span>
{x-2y=6 => 2y=x-6 => y=(x-6)/2 => y=x/2-3
{f(x)=6/(x-2)
{f(x)=x/2-3
Графическое решение неравенства: (0;-3) => x</span>₁=0; y₁=-3
(8;1) => x₂=8; y₂=1
Проверка: {(x-2)y=6 => y=6/(x-2) ОДЗ: x≠2
{x-2y=6
x-2(6/(x-2)=6
x(x-2)-12=6(x-2)
x²-2x-12-6x+12=0
x²-8x=0
x(x-8)=0
x=0 или x-8=0
x₁=0; x₂=8
y=6/(-2) или y=6/(8-2)
y₁=-3; y₂=1
График во вложении